To solve the integral \( \int_1^2 |2x - [3x]| \, dx \), we need to analyze the expression inside the absolute value. Here’s a step-by-step breakdown of the solution:
### Step 1: Analyze the expression \( |2x - [3x]| \)
The term \( [3x] \) is the greatest integer function (or floor function) of \( 3x \). We need to determine how \( [3x] \) behaves in the interval from \( x = 1 \) to \( x = 2 \).
- For \( x = 1 \): \( 3x = 3 \) and \( [3] = 3 \)
- For \( x = 2 \): \( 3x = 6 \) and \( [6] = 6 \)
Now, we can find the values of \( [3x] \) in the interval \( [1, 2] \):
- For \( 1 \leq x < \frac{4}{3} \) (i.e., \( 1 \leq 3x < 4 \)): \( [3x] = 3 \)
- For \( \frac{4}{3} \leq x < \frac{5}{3} \) (i.e., \( 4 \leq 3x < 5 \)): \( [3x] = 4 \)
- For \( \frac{5}{3} \leq x < 2 \) (i.e., \( 5 \leq 3x < 6 \)): \( [3x] = 5 \)
### Step 2: Break the integral into intervals
Now, we can break the integral into three parts based on the intervals we found:
1. From \( x = 1 \) to \( x = \frac{4}{3} \)
2. From \( x = \frac{4}{3} \) to \( x = \frac{5}{3} \)
3. From \( x = \frac{5}{3} \) to \( x = 2 \)
### Step 3: Evaluate the integral in each interval
**Interval 1: \( [1, \frac{4}{3}) \)**
In this interval, \( [3x] = 3 \):
\[
|2x - [3x]| = |2x - 3| = 3 - 2x \quad (\text{since } 2x < 3)
\]
Thus, the integral becomes:
\[
\int_1^{\frac{4}{3}} (3 - 2x) \, dx
\]
Calculating this integral:
\[
= \left[ 3x - x^2 \right]_1^{\frac{4}{3}} = \left( 3 \cdot \frac{4}{3} - \left( \frac{4}{3} \right)^2 \right) - \left( 3 \cdot 1 - 1^2 \right)
\]
\[
= \left( 4 - \frac{16}{9} \right) - (3 - 1) = \left( \frac{36}{9} - \frac{16}{9} \right) - 2 = \frac{20}{9} - 2 = \frac{20}{9} - \frac{18}{9} = \frac{2}{9}
\]
**Interval 2: \( [\frac{4}{3}, \frac{5}{3}) \)**
In this interval, \( [3x] = 4 \):
\[
|2x - [3x]| = |2x - 4| = 2x - 4 \quad (\text{since } 2x > 4)
\]
Thus, the integral becomes:
\[
\int_{\frac{4}{3}}^{\frac{5}{3}} (2x - 4) \, dx
\]
Calculating this integral:
\[
= \left[ x^2 - 4x \right]_{\frac{4}{3}}^{\frac{5}{3}} = \left( \left( \frac{5}{3} \right)^2 - 4 \cdot \frac{5}{3} \right) - \left( \left( \frac{4}{3} \right)^2 - 4 \cdot \frac{4}{3} \right)
\]
\[
= \left( \frac{25}{9} - \frac{20}{3} \right) - \left( \frac{16}{9} - \frac{16}{3} \right)
\]
\[
= \left( \frac{25}{9} - \frac{60}{9} \right) - \left( \frac{16}{9} - \frac{48}{9} \right) = \left( -\frac{35}{9} \right) - \left( -\frac{32}{9} \right) = -\frac{35}{9} + \frac{32}{9} = -\frac{3}{9} = -\frac{1}{3}
\]
**Interval 3: \( [\frac{5}{3}, 2] \)**
In this interval, \( [3x] = 5 \):
\[
|2x - [3x]| = |2x - 5| = 5 - 2x \quad (\text{since } 2x < 5)
\]
Thus, the integral becomes:
\[
\int_{\frac{5}{3}}^{2} (5 - 2x) \, dx
\]
Calculating this integral:
\[
= \left[ 5x - x^2 \right]_{\frac{5}{3}}^{2} = \left( 5 \cdot 2 - 2^2 \right) - \left( 5 \cdot \frac{5}{3} - \left( \frac{5}{3} \right)^2 \right)
\]
\[
= (10 - 4) - \left( \frac{25}{3} - \frac{25}{9} \right) = 6 - \left( \frac{75}{9} - \frac{25}{9} \right) = 6 - \frac{50}{9} = \frac{54}{9} - \frac{50}{9} = \frac{4}{9}
\]
### Step 4: Combine the results
Now, we sum up the results from all three intervals:
\[
\text{Total} = \frac{2}{9} - \frac{1}{3} + \frac{4}{9}
\]
Converting \( -\frac{1}{3} \) to ninths:
\[
-\frac{1}{3} = -\frac{3}{9}
\]
Thus,
\[
\text{Total} = \frac{2}{9} - \frac{3}{9} + \frac{4}{9} = \frac{2 - 3 + 4}{9} = \frac{3}{9} = \frac{1}{3}
\]
### Final Answer
The value of the integral \( \int_1^2 |2x - [3x]| \, dx = \frac{1}{3} \).
